Divide the following complex numbers. $ \dfrac{-23+11i}{5+i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${5-i}$ $ \dfrac{-23+11i}{5+i} = \dfrac{-23+11i}{5+i} \cdot \dfrac{{5-i}}{{5-i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-23+11i) \cdot (5-i)} {(5+i) \cdot (5-i)} = \dfrac{(-23+11i) \cdot (5-i)} {5^2 - (1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-23+11i) \cdot (5-i)} {(5)^2 - (1i)^2} = $ $ \dfrac{(-23+11i) \cdot (5-i)} {25 + 1} = $ $ \dfrac{(-23+11i) \cdot (5-i)} {26} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-23+11i}) \cdot ({5-i})} {26} = $ $ \dfrac{{-23} \cdot {5} + {11} \cdot {5 i} + {-23} \cdot {-1 i} + {11} \cdot {-1 i^2}} {26} $ Evaluate each product of two numbers. $ \dfrac{-115 + 55i + 23i - 11 i^2} {26} $ Finally, simplify the fraction. $ \dfrac{-115 + 55i + 23i + 11} {26} = \dfrac{-104 + 78i} {26} = -4+3i $